Lösung 3.1:4d

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In the equation, <math>z</math> occurs only as <math>\bar{z}</math> and, to begin with, we can therefore treat <math>\bar{z}</math> as unknown.
In the equation, <math>z</math> occurs only as <math>\bar{z}</math> and, to begin with, we can therefore treat <math>\bar{z}</math> as unknown.
Divide both sides by <math>2+i</math>,
Divide both sides by <math>2+i</math>,
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{{Displayed math||<math>\bar{z}=\frac{1+i}{2+i}\,,</math>}}
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<math>\bar{z}=\frac{1+i}{2+i}</math>
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and calculate the quotient on the right-hand side by multiplying top and bottom by the complex conjugate of the denominator,
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{{Displayed math||<math>\begin{align}
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\bar{z}
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&= \frac{(1+i)(2-i)}{(2+i)(2-i)}
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= \frac{1\cdot 2-1\cdot i +i \cdot 2 - i\cdot i}{2^2-i^2}\\[5pt]
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&= \frac{2-i+2i+1}{4+1}
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= \frac{3+i}{5}
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= \frac{3}{5}+\frac{1}{5}\,i\,\textrm{.}
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\end{align}</math>}}
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and calculate the quotient on the right-hand side by multiplying top and bottom by the complex conjugate of the numerator:
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This means that <math>z=\tfrac{3}{5}-\tfrac{1}{5}i\,</math>.
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We check that <math>z=\tfrac{3}{5}-\tfrac{1}{5}i</math> satisfies the original equation,
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<math>\begin{align}\bar{z}&=\frac{(1+i)(2-i)}{(2+i)(2-i)}=\frac{1\cdot 2-1\cdot i +i \cdot 2 - i\cdot i}{2^2-i^2}\\
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{{Displayed math||<math>\begin{align}
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&=\frac{2-i+2i+1}{4+1}=\frac{3+i}{5}=\frac{3}{5}+\frac{1}{5}i.\end{align}</math>
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\text{LHS}
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&= (2+i)\bar{z}
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= (2+i)\overline{\Bigl(\frac{3}{5}-\frac{1}{5}\,i\Bigr)}
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This means that <math>z=\frac{3}{5}-\frac{1}{5}i.</math>
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= (2+i)\Bigl(\frac{3}{5}+\frac{1}{5}\,i\Bigr)\\[5pt]
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&= 2\cdot\frac{3}{5}+2\cdot\frac{1}{5}\,i+i\cdot\frac{3}{5}+i\cdot\frac{1}{5}\,i
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We check that <math>z=\frac{3}{5}-\frac{1}{5}i</math> satisfies the original equation:
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= \frac{6}{5}+\frac{2}{5}\,i+\frac{3}{5}\,i-\frac{1}{5}\\[5pt]
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&=\frac{6-1}{5}+\frac{2+3}{5}\,i
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= 1+i
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<math>\begin{align}LHS &= (2+i)\bar{z} = (2+i)\overline{(\frac{3}{5}-\frac{1}{5}i)}=(2+i)(\frac{3}{5}+\frac{1}{5}i)\\
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= \text{RHS}\,\textrm{.}
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&=2\cdot\frac{3}{5}+2\cdot\frac{1}{5}i+i\cdot\frac{3}{5}+i\cdot\frac{1}{5}i=\frac{6}{5}+\frac{2}{5}i+\frac{3}{5}i-\frac{1}{5}\\
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\end{align}</math>}}
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&=\frac{6-1}{5}+\frac{2+3}{5}i=1+i= RHS\end{align}</math>
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Version vom 08:09, 30. Okt. 2008

In the equation, \displaystyle z occurs only as \displaystyle \bar{z} and, to begin with, we can therefore treat \displaystyle \bar{z} as unknown.

Divide both sides by \displaystyle 2+i,

\displaystyle \bar{z}=\frac{1+i}{2+i}\,,

and calculate the quotient on the right-hand side by multiplying top and bottom by the complex conjugate of the denominator,

\displaystyle \begin{align}

\bar{z} &= \frac{(1+i)(2-i)}{(2+i)(2-i)} = \frac{1\cdot 2-1\cdot i +i \cdot 2 - i\cdot i}{2^2-i^2}\\[5pt] &= \frac{2-i+2i+1}{4+1} = \frac{3+i}{5} = \frac{3}{5}+\frac{1}{5}\,i\,\textrm{.} \end{align}

This means that \displaystyle z=\tfrac{3}{5}-\tfrac{1}{5}i\,.

We check that \displaystyle z=\tfrac{3}{5}-\tfrac{1}{5}i satisfies the original equation,

\displaystyle \begin{align}

\text{LHS} &= (2+i)\bar{z} = (2+i)\overline{\Bigl(\frac{3}{5}-\frac{1}{5}\,i\Bigr)} = (2+i)\Bigl(\frac{3}{5}+\frac{1}{5}\,i\Bigr)\\[5pt] &= 2\cdot\frac{3}{5}+2\cdot\frac{1}{5}\,i+i\cdot\frac{3}{5}+i\cdot\frac{1}{5}\,i = \frac{6}{5}+\frac{2}{5}\,i+\frac{3}{5}\,i-\frac{1}{5}\\[5pt] &=\frac{6-1}{5}+\frac{2+3}{5}\,i = 1+i = \text{RHS}\,\textrm{.} \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.1:4d“